redox - How do metals split the oxygen double bond during corrosion?

For a metal like $\ce{Mg}$ to corrode into $\ce{MgO}$, the double bond in $\ce{O2}$ must break. Since the dissociation energy of $\ce{O2}$ is 500 kJ/mol, using estimates from the Boltzmann distribution, it looks like there would not be any $\ce{O2}$ molecules in the entire atmosphere with enough energy to break this bond at room temperature: $$\text{Fraction of $\ce{O2}$ with at least $\pu{500 kJ}$ energy} \approx \mathrm e^\left(\dfrac{-500\,000\ \mathrm{J/mol}}{298\ \mathrm K(8.3145\ \mathrm{J/(mol\ K)})}\right) \lt 10^{-261}.$$

But my understanding is that $\ce{MgO}$ does form at room temperature. In fact, it seems that even stable $\ce{Pt}$ can break the $\ce{O2}$ bond at room temperature (for example in the context of catalyzing the ignition of hydrogen).

It seems that something like this must be happening: an $\ce{O2}$ molecule collides with $\ce{Mg}$ metal and gets stuck onto the surface, temporarily forming $\ce{MgO2}$. And once this bond is formed, the remaining bond between the two oxygen atoms suddenly becomes easy to break, and one of the oxygen atoms falls off and bonds with a different $\ce{Mg}$ atom. Is this the case?

I'd like to understand this mechanism better. How weak does the $\ce{O2}$ bond become after latching onto magnesium, and why does it become weaker? Why are metals able to split the $\ce{O2}$ bond this way, compared to other materials like hydrocarbons?